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Math Help - Second order non homogeneous DE, annihilator method

  1. #1
    MHF Contributor arbolis's Avatar
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    Second order non homogeneous DE, annihilator method

    I've followed Chris' tutorial, but I didn't get lucky at the end.
    I must solve y''+2y'=xe^{-2x}. Also, y(0)=a and y'(0)=b.
    So I consider the homogeneous equation y''+2y'=0. I wrote the characteristic polynomial and found its roots to be \pm i \sqrt 2. Thus a solution to the homogeneous DE is c_1 \cos (x\sqrt 2)+c_2 \sin (x \sqrt 2). I'm not sure if I can solve for c_1 and c_2 yet, with the initial conditions. It would be the solution the homogeneous DE, but are these constants the same in the case of the non homogeneous case? I guess yes...
    Anyway, I continued writing (D^2+2D)(y)=xe^{-2x}. The annihilator is (D+2).
    Hence (D+2)(D^2+2D)(y)=0. So (r+2)(r^2+2r)=0 \Rightarrow r=0 or r=-2 (this one is a double root).
    Thus a general solution to the original DE is y=c_1 \cos (x\sqrt 2)+c_2 \sin (x \sqrt 2)+c_3e^{-2x}+c_4.
    I tried to solve for c_3 and c_4 this way:
    y_p=c_3e^{-2x}.
    y_p'=-2c_3e^{-2x}.
    y_p''=4c_3e^{-2x}.
    Plugging these values into the original DE, I reach that xe^{-2x}=0, c_3 cancels out... Obviously I made something wrong. I don't see what.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arbolis View Post
    I've followed Chris' tutorial, but I didn't get lucky at the end.
    I must solve y''+2y'=xe^{-2x}. Also, y(0)=a and y'(0)=b.
    So I consider the homogeneous equation y''+2y'=0. I wrote the characteristic polynomial and found its roots to be \pm i \sqrt 2. Thus a solution to the homogeneous DE is c_1 \cos (x\sqrt 2)+c_2 \sin (x \sqrt 2). I'm not sure if I can solve for c_1 and c_2 yet, with the initial conditions. It would be the solution the homogeneous DE, but are these constants the same in the case of the non homogeneous case? I guess yes...
    Anyway, I continued writing (D^2+2D)(y)=xe^{-2x}. The annihilator is (D+2).
    Hence (D+2)(D^2+2D)(y)=0. So (r+2)(r^2+2r)=0 \Rightarrow r=0 or r=-2 (this one is a double root).
    Thus a general solution to the original DE is y=c_1 \cos (x\sqrt 2)+c_2 \sin (x \sqrt 2)+c_3e^{-2x}+c_4.
    I tried to solve for c_3 and c_4 this way:
    y_p=c_3e^{-2x}.
    y_p'=-2c_3e^{-2x}.
    y_p''=4c_3e^{-2x}.
    Plugging these values into the original DE, I reach that xe^{-2x}=0, c_3 cancels out... Obviously I made something wrong. I don't see what.
    If you consider the homogeneous equation y^{\prime\prime}+2y^{\prime}=0, the associated characteristic equation is r^2+2r=0\implies r_1=0 and r_2=-2.

    So the homogeneous solution is y=c_1+ c_2e^{-2x}

    Now, we consider the non-homogeneous equation in differential operator form:

    (D^2+2D)y=xe^{-2x}.

    Since \left(D-\alpha\right)^n is the annihilator of e^{\alpha x},xe^{\alpha x},\ldots, x^{n-1}e^{\alpha x}, it follows that (D+2)^2 annihilates xe^{-2x}.

    Therefore, when you apply the annihilator to both sides, we get

    (D+2)^2(D^2+2D)(y)=0.

    The characteristic equation is (r+2)^2({\color{red}r^2+2r})=0

    The part in red generates the complimentary solution. What we're interested in is the (r+2)^2 term. This generates solutions r=-2 with multiplicity 2 (but don't forget r+2 was a term in the complementary case!).

    This implies that the particular solution will be of the form y_p=Axe^{-2x}+Bx^2e^{-2x}.

    Now, substitute this into your original differential equation to solve for the constants A and B.

    Once you get the particular solution, the general solution will be of the form y=c_1 + c_2e^{-2x}+y_p(x).

    It is at this point that you apply the initial conditions that you were given at the start of the problem.

    Can you take it from here?
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thank you very much Chris!
    I don't know why I considered the equation y''+2y=0 (hence the roots I found!) instead of y''+2y'=0. I'll try to do the rest alone, but in any case I'm going to refer to your help in this post.
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  4. #4
    MHF Contributor arbolis's Avatar
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    I've reached a result, but I will re do the arithmetics tomorrow (it's 1:30 am and I'm so tired...).
    I just don't know why
    This implies that the particular solution will be of the form y_p=Axe^{-2x}+Bx^2e^{-2x}
    . Why the " x" and " x^2" terms? I will read the theory for sure, but if you could kill this doubt I have, I'd be grateful. Of course you can refer me to your DE tutorial.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arbolis View Post
    I've reached a result, but I will re do the arithmetics tomorrow (it's 1:30 am and I'm so tired...).
    I just don't know why . Why the " x" and " x^2" terms? I will read the theory for sure, but if you could kill this doubt I have, I'd be grateful. Of course you can refer me to your DE tutorial.
    The reason for that is because you already have an e^{-2x} in the homogeneous solution. So by repeated roots, the functions in your particular solution are xe^{-2x} and x^2e^{-2x}.

    Does this clarify things?
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  6. #6
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    The reason for that is because you already have an e^{-2x} in the homogeneous solution. So by repeated roots, the functions in your particular solution are xe^{-2x} and x^2e^{-2x}.

    Does this clarify things?
    Oh... yes it does. I've just read the example 18 of your tutorial and now things are perfectly clear. Once again, thanks a bunch.


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